Question

The sum of the coefficients of \( x^{2/3} \) and \( x^{-2/5} \) in the binomial expansion of $$ \left( x^{2/3} + \frac{1}{2} x^{-2/5} \right)^9 $$ is: 

• A. \( \frac{21}{4} \)
• B. \( \frac{69}{16} \)
• C. \( \frac{63}{16} \)
• D. \( \frac{19}{4} \)

Answer 1

The Correct Option is A

Step 1. General Term of the Expansion:
The general term in the binomial expansion of \( \left( x^{2/3} + \frac{1}{2}x^{-2/5} \right)^9 \) is given by:

\[ T_{r+1} = \binom{9}{r} \left( x^{2/3} \right)^{9-r} \left( \frac{x^{-2/5}}{2} \right)^r \]

Simplify the expression:

\[ T_{r+1} = \binom{9}{r} \left( \frac{1}{2} \right)^r x^{\left( \frac{6 - 2r}{3} - \frac{2r}{5} \right)} \]

Step 2. For \( x^{2/3} \):
Set the power of \( x \) equal to \( 2/3 \):

\[ \frac{6 - 2r}{3} - \frac{2r}{5} = \frac{2}{3} \]

Solving this equation gives \( r = 5 \).
Substituting \( r = 5 \) into the coefficient formula:

\[ \text{Coefficient of } x^{2/3} = \binom{9}{5} \left( \frac{1}{2} \right)^5 \]

Step 3. For \( x^{-2/5} \):
Set the power of \( x \) equal to \( -2/5 \):

\[ \frac{6 - 2r}{3} - \frac{2r}{5} = -\frac{2}{5} \]

Solving this equation gives \( r = 6 \).
Substituting \( r = 6 \) into the coefficient formula:

\[ \text{Coefficient of } x^{-2/5} = \binom{9}{6} \left( \frac{1}{2} \right)^6 \]

Step 4. Sum of the Coefficients:
Add the two coefficients:

\[ \text{Sum} = \binom{9}{5} \left( \frac{1}{2} \right)^5 + \binom{9}{6} \left( \frac{1}{2} \right)^6 \]

Simplify:\[ \text{Sum} = \frac{21}{4}\]